The ratio of potential difference that must b | vedclass

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(B) Given: $C_1: C_2 = 1: 2$. Therefore,$C_2 = 2C_1$.Equivalent capacitance in parallel: $C_P = C_1 + C_2 = C_1 + 2C_1 = 3C_1$.Equivalent capacitance

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$\sqrt{2}:3$

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(B) Given: $C_1: C_2 = 1: 2$. Therefore,$C_2 = 2C_1$.Equivalent capacitance in parallel: $C_P = C_1 + C_2 = C_1 + 2C_1 = 3C_1$.Equivalent capacitance in series: $C_S = \frac{C_1 C_2}{C_1 + C_2} = \frac{C_1(2C_1)}{3C_1} = \frac{2}{3}C_1$.Let $V_P$ and $V_S$ be the potential differences applied across the parallel and series combinations respectively.Since the energy stored $E = \frac{1}{2}CV^2$ is the same in both cases:$\frac{1}{2} C_P V_P^2 = \frac{1}{2} C_S V_S^2$$\frac{V_P^2}{V_S^2} = \frac{C_S}{C_P} = \frac{\frac{2}{3}C_1}{3C_1} = \frac{2}{9}$Taking the square root on both sides:$\frac{V_P}{V_S} = \sqrt{\frac{2}{9}} = \frac{\sqrt{2}}{3}$.

The ratio of potential difference that must be applied across parallel and series combination of two capacitors $C_1$ and $C_2$ with their capacitance in the ratio $1: 2$ so that energy stored in these two cases becomes same is

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